\(\begin{array}{l}
a)\\
Phan\,1:\\
C_aH_{2a+1}OH+\frac{3a}{2}O_2\xrightarrow{t^o}aCO_2+(a+1)H_2O\\
C_bH_{2b+1}COOH+\frac{3b+1}{2}O_2\xrightarrow{t^o}(b+1)CO_2+(b+1)H_2O\\
CO_2+Ca(OH)_2\to CaCO_3+H_2O(1)\\
2CO_2+Ca(OH)_2\to Ca(HCO_3)_2\\
Ca(HCO_3)_2\xrightarrow{t^o}CaCO_3+CO_2+H_2O(2)\\
Phan\,2:\\
C_bH_{2b+1}COOH+NaOH\to C_bH_{2b+1}COONa+H_2O\\
Phan\,3:\\
C_aH_{2a+1}OH+C_bH_{2b+1}COOH\buildrel{{t^o,H_2SO_{4(đ)}}}\over\rightleftharpoons C_bH_{2b+1}COOC_aH_{2a+1}+H_2O\\
b)\\
Moi\,phan\,dat\;n_{C_bH_{2b+1}COOH}=x(mol);n_{C_aH_{2a+1}OH}=y(mol)\\
n_{CaCO_3(1)}=\frac{30}{100}=0,3(mol)\\
n_{CaCO_3(2)}=\frac{10}{100}=0,1(mol)\\
n_{Ca(HCO_3)_2}=n_{CaCO_3(2)}=0,1(mol)\\
\sum n_{CO_2}=n_{CaCO_3(1)}+2n_{Ca(HCO_3)_2}=0,5(mol)\\
m_{bình\,tăng}=m_{CO_2}+m_{H_2O}\\
\to m_{H_2O}=34,6-0,5.44=12,6(g)\\
\to n_{H_2O}=0,7(mol)\\
Theo\,PT:x=n_{C_bH_{2b+1}COOH}=n_{NaOH}=0,5.0,1=0,05(mol)\\
Theo\,PT:\begin{cases} n_{CO_2}=(b+1)n_{C_bH_{2b+1}COOH}+a.n_{C_aH_{2n+1}OH}\\ n_{H_2O}=(a+1)n_{C_aH_{2a+1}OH}+(b+1)n_{C_bH_{2b+1}OH}\end{cases}\\
\to \begin{cases} (b+1).0,05+ay=0,5\\ (a+1)y+(b+1).0,05=0,7\\ a=b+1 \end{cases}\\
\to \begin{cases} 0,05b+ay=0,45\\ ay+0,05b+y=0,65\\ a=b+1 \end{cases}\\
\to\begin{cases} y=0,2\\ 0,05b+0,2a=0,45\\ a=b+1 \end{cases}\\
\to\begin{cases} a=2\\ b=1 \end{cases}\\
\to \begin{cases} X:C_2H_5OH\\ Y:CH_3COOH \end{cases}\\
c)\\
n_{C_2H_5OH}=0,2(mol);n_{CH_3COOH}=0,05(mol)\\
C_2H_5OH+CH_3COOH\buildrel{{t^o,H_2SO_{4(đ)}}}\over\rightleftharpoons CH_3COOC_2H_5+H_2O\\
Vi\,n_{C_2H_5OH}>n_{CH_3COOH}\to Hieu\,suat\,tinh\,theo\,CH_3COOH\\
n_{CH_3COOC_2H_5(lt)}=n_{CH_3COOH(lt)}=0,05(mol)\\
m=m_{CH_3COOC_2H_5(tt)}=0,05.88.75\%=3,3(g)
\end{array}\)

 `a)` 

`C_aH_{2a+1}OH + {3a}/2O_2 → aCO_2 + (a+1)H_2O` 

`C_bH_{2b+1}COOH + {3b+1}/2O_2→ (b + 1)CO_2 + (b+1)H_2O`

`CO_2 + Ca(OH)_2 → CaCO_3+H_2O` 

`CaCO_3 + CO_2 + H_2O → Ca(HCO_3)_2`

`Ca(HCO_3)_2 → CaCO_3 + CO_2 + H_2O`

`C_bH_{2b+1}COOH + NaOH → C_bH_{2b+1}COONa + H_2O` 

`C_aH_{2a+1}OH + C_bH_{2b + 1}COOH → C_bH_{2b+1}COOC_aH_{2a+1} + H_2O` 

`b)`

`n_{CaCO_3} = {30}/{100} = 0,3(mol)` 

`n_{CaCO_3}(HCO_3^{-}) = {10}/{100} = 0,1(mol)` 

Theo pt: 

`n_{CO_2} = n_{Ca(HCO_3)_2} = 0,1(mol)`

`n_{CO_2} = n_{Ca(HCO_3)_2} + n_{CaCO_3} + n_{CO_2} `

`= 0,3 + 0,1+0,1 = 0,5(mol)` 

`m↑ = m_{CO_2} + m_{H_2O}`

`<=> 34,6 = 0,5 .44 + m_{H_2O}`

`→ m_{H_2O} = 12,6(g)`

`n_{H_2O} = {12,6}/{18} = 0,7(mol)`

`n_{NaOH} = {100}/{1000} . 0,5 = 0,05(mol)` 

Theo pt: 

`n_{text{axit}} = n_{NaOH} = 0,05(mol)` 

Mỗi phần:

`n_{C_aH_{2a+1}OH} = x(mol)`

`n_{C_bH_{2b+1}COOH = 0,05(mol)`

Từ dữ kiện và phương trình: 

$\begin{cases} n_{CO_2} = ax + 0,05(b+1) = 0,5(mol)\\n_{H_2O} = x(a+1) + 0,05(b+1) = 0,7(mol)\end{cases}$

`a = b + 1` 

`→ `$\begin{cases} a = 2\\b = 1\\y = 0,2(mol) \end{cases}$

`→` $\begin{cases} C_aH_{2a+1}OH: C_2H_5OH\\C_bH_{2b+1}COOH: CH_3COOH \end{cases}$

`c)`

`C_2H_5OH + CH_3COOH → CH_3COOC_2H_5 + H_2O`

`n_{C_2H_5OH}>n_{CH_3COOH}` 

`→ n_{text{este}} = n_{CH_3COOH} = 0,05(mol)`

`m = 0,05 . 88 . 75% = 3,3(g)`